Let the first three terms 2, p and q, which `q ne 2`, of a G.P. be respectively the 7th, 8th and 13th terms of an A.P. If the 5th term of the G.P. is the nth terms of the A.P. then n is equal to

Options

A. 163
B. 177
C. 151
D. 169

Answer

To solve the problem, we need to establish the relationships between the terms of the geometric progression (G.P.) and the arithmetic progression (A.P.) as given in the question.

### Step-by-Step Solution:

1. **Identify the Terms of G.P.**:
The first three terms of the G.P. are given as \(2\), \(p\), and \(q\). In a G.P., if the first term is \(a\) and the common ratio is \(r\), the terms can be expressed as:
- First term: \(a = 2\)
- Second term: \(p = ar = 2r\)
- Third term: \(q = ar^2 = 2r^2\)

2. **Identify the Terms of A.P.**:
The terms \(2\), \(p\), and \(q\) are also the 7th, 8th, and 13th terms of an A.P. respectively. The \(n\)th term of an A.P. can be expressed as:
\[ A_n = A + (n-1)D \]
where \(A\) is the first term and \(D\) is the common difference.

3. **Set Up Equations for A.P.**:
From the A.P. definitions:
- For the 7th term: \(A + 6D = 2\) (1)
- For the 8th term: \(A + 7D = p\) (2)
- For the 13th term: \(A + 12D = q\) (3)

4. **Express \(p\) and \(q\) in Terms of \(A\) and \(D\)**:
From equations (1) and (2):
\[ p - 2 = (A + 7D) - (A + 6D) = D \quad \Rightarrow \quad D = p - 2 \quad (4) \]
From equations (1) and (3):
\[ q - 2 = (A + 12D) - (A + 6D) = 6D \quad \Rightarrow \quad q = 2 + 6D \quad (5) \]

5. **Substitute \(D\) from (4) into (5)**:
Substitute \(D = p - 2\) into \(q = 2 + 6D\):
\[ q = 2 + 6(p - 2) = 2 + 6p - 12 = 6p - 10 \quad (6) \]

6. **Relate \(p\) and \(q\) from G.P.**:
From G.P., we have \(p = 2r\) and \(q = 2r^2\). Substitute these into equation (6):
\[ 2r^2 = 6(2r) - 10 \quad \Rightarrow \quad 2r^2 = 12r - 10 \]
Rearranging gives:
\[ 2r^2 - 12r + 10 = 0 \quad \Rightarrow \quad r^2 - 6r + 5 = 0 \]

7. **Solve the Quadratic Equation**:
Factor the quadratic:
\[ (r - 1)(r - 5) = 0 \]
Thus, \(r = 1\) or \(r = 5\). Since \(q \neq 2\), we discard \(r = 1\) and take \(r = 5\).

8. **Calculate \(p\) and \(q\)**:
Substitute \(r = 5\):
\[ p = 2r = 2 \times 5 = 10 \]
\[ q = 2r^2 = 2 \times 25 = 50 \]

9. **Find the 5th Term of the G.P.**:
The 5th term of the G.P. is given by:
\[ A_5 = ar^4 = 2 \times 5^4 = 2 \times 625 = 1250 \]

10. **Set Up Equation for the nth Term of A.P.**:
We know that the 5th term of the G.P. is the nth term of the A.P.:
\[ 1250 = A + (n-1)D \]
From (1), we have \(A + 6D = 2\), thus:
\[ A = 2 - 6D \]
Substitute \(D = p - 2 = 10 - 2 = 8\):
\[ A = 2 - 6 \times 8 = 2 - 48 = -46 \]

11. **Substitute \(A\) and \(D\) into the nth Term Equation**:
Substitute \(A\) and \(D\):
\[ 1250 = -46 + (n-1) \times 8 \]
Rearranging gives:
\[ 1250 + 46 = (n-1) \times 8 \quad \Rightarrow \quad 1296 = (n-1) \times 8 \]
\[ n - 1 = \frac{1296}{8} = 162 \quad \Rightarrow \quad n = 163 \]

### Final Answer:
The value of \(n\) is \(163\).